NO
proof of /export/starexec/sandbox2/benchmark/theBenchmark.c
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


Termination of the given C Problem could be disproven:

(0) C Problem
(1) CToIRSProof [EQUIVALENT, 0 ms]
(2) IntTRS
(3) IRS2T2 [EQUIVALENT, 0 ms]
(4) T2IntSys
(5) T2 [COMPLETE, 1393 ms]
(6) NO


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(0)
Obligation:
c file /export/starexec/sandbox2/benchmark/theBenchmark.c
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(1) CToIRSProof (EQUIVALENT)
Parsed C Integer Program as IRS.
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(2)
Obligation:
Rules:
f1(i) -> f2(x_1) :|: TRUE
f7(x) -> f10(arith) :|: TRUE && arith = x + 1
f4(x1) -> f7(x1) :|: x1 < 0
f4(x2) -> f8(x2) :|: x2 >= 0
f10(x3) -> f9(x3) :|: TRUE
f8(x4) -> f9(x4) :|: TRUE
f11(x17) -> f14(x18) :|: TRUE && x18 = x17 - 1
f9(x6) -> f11(x6) :|: x6 > 0
f9(x7) -> f12(x7) :|: x7 <= 0
f14(x8) -> f13(x8) :|: TRUE
f12(x9) -> f13(x9) :|: TRUE
f3(x10) -> f4(x10) :|: x10 > 0 - 5 && x10 < 5
f3(x11) -> f5(x11) :|: x11 <= 0 - 5
f3(x19) -> f5(x19) :|: x19 >= 5
f13(x12) -> f6(x12) :|: TRUE
f5(x13) -> f6(x13) :|: TRUE
f2(x14) -> f3(x14) :|: x14 < 0
f2(x20) -> f3(x20) :|: x20 > 0
f6(x15) -> f2(x15) :|: TRUE
f2(x16) -> f15(x16) :|: x16 = 0
Start term: f1(i)

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(3) IRS2T2 (EQUIVALENT)
Transformed input IRS into an integer transition system.Used the following mapping from defined symbols to location IDs:

   (f1_1,1)
   (f2_1,2)
   (f7_1,3)
   (f10_1,4)
   (f4_1,5)
   (f8_1,6)
   (f9_1,7)
   (f11_1,8)
   (f14_1,9)
   (f12_1,10)
   (f13_1,11)
   (f3_1,12)
   (f5_1,13)
   (f6_1,14)
   (f15_1,15)

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(4)
Obligation:
START: 1;

FROM: 1;
oldX0 := x0;
oldX1 := nondet();
assume(0 = 0);
x0 := oldX1;
TO: 2;

FROM: 3;
oldX0 := x0;
oldX1 := -(-(oldX0 + 1));
assume(0 = 0 && oldX1 = oldX0 + 1);
x0 := -(-(oldX0 + 1));
TO: 4;

FROM: 5;
oldX0 := x0;
assume(oldX0 < 0);
x0 := oldX0;
TO: 3;

FROM: 5;
oldX0 := x0;
assume(oldX0 >= 0);
x0 := oldX0;
TO: 6;

FROM: 4;
oldX0 := x0;
assume(0 = 0);
x0 := oldX0;
TO: 7;

FROM: 6;
oldX0 := x0;
assume(0 = 0);
x0 := oldX0;
TO: 7;

FROM: 8;
oldX0 := x0;
oldX1 := -(1 - oldX0);
assume(0 = 0 && oldX1 = oldX0 - 1);
x0 := -(1 - oldX0);
TO: 9;

FROM: 7;
oldX0 := x0;
assume(oldX0 > 0);
x0 := oldX0;
TO: 8;

FROM: 7;
oldX0 := x0;
assume(oldX0 <= 0);
x0 := oldX0;
TO: 10;

FROM: 9;
oldX0 := x0;
assume(0 = 0);
x0 := oldX0;
TO: 11;

FROM: 10;
oldX0 := x0;
assume(0 = 0);
x0 := oldX0;
TO: 11;

FROM: 12;
oldX0 := x0;
assume(oldX0 > 0 - 5 && oldX0 < 5);
x0 := oldX0;
TO: 5;

FROM: 12;
oldX0 := x0;
assume(oldX0 <= 0 - 5);
x0 := oldX0;
TO: 13;

FROM: 12;
oldX0 := x0;
assume(oldX0 >= 5);
x0 := oldX0;
TO: 13;

FROM: 11;
oldX0 := x0;
assume(0 = 0);
x0 := oldX0;
TO: 14;

FROM: 13;
oldX0 := x0;
assume(0 = 0);
x0 := oldX0;
TO: 14;

FROM: 2;
oldX0 := x0;
assume(oldX0 < 0);
x0 := oldX0;
TO: 12;

FROM: 2;
oldX0 := x0;
assume(oldX0 > 0);
x0 := oldX0;
TO: 12;

FROM: 14;
oldX0 := x0;
assume(0 = 0);
x0 := oldX0;
TO: 2;

FROM: 2;
oldX0 := x0;
assume(oldX0 = 0);
x0 := oldX0;
TO: 15;


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(5) T2 (COMPLETE)
Found this recurrent set for cutpoint 12: oldX1 == 5 and x0 == 5

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(6)
NO